### Solving the quartic using “obvious” steps

Looking over the derivations of solutions to quartics (fourth-degree polynomials), I was a little disturbed at how, except for René Descartes’s, the derivations all have steps where “magic” occurs, by which I mean that a step does something mathematically valid but very unintuitive, which just happens to make stuff later on work out just right.

So I wondered whether one could take an approach like Lodovico Ferrari’s, but do it without his “magic”. The basic idea to Ferrari’s approach is to convert the depressed^{*} quartic equation *f*(*x*) = *x*⁴ + *αx*² + *βx* + *γ* = 0 into (*x*²+*p*)² = *m*(*x*+*q*)², because taking the square root of both sides produces the quadratic equation *x*²+*p* = ±(*x*+*q*)√*m*.

(*x*²+*p*)² = *m*(*x*+*q*)² expands into *x*⁴ + 2*px*² + *p*² = *m*(*x*²+2*xq*+*q*²), which is *x*⁴ + (2*p−m*)*x*² − 2*mqx* + *p*² − *mq*² = 0. We can name that polynomial *g*(*x*). From above, we know that the roots of *g* are the same as the roots of *x*² ± (*x*+*q*)√*m* + *p*, and the quadratic formula easily gives us those roots in terms of *m*, *p*, & *q*.

*g* ≡ *f* iff *α* = 2*p−m* and *β* = −2*mq* and *γ = p*²−*mq*². Thus our task is to get values for *p*, *q*, & *m* which satisfy these three equations.

*m* = −*β*/2*q* and *p* = ½(*α*+*m*), so *p* = ½(*α* − *β*/2*q*).

*γ* = *p*² − *q*²*m* = *p*² + *q*²*β*/2*q* = *p*² + ½*βq* = [½(*α − β*/2*q*)]² + ½*βq* = ¼[*α*² − *αβ*/*q* + *β* ²/4*q*² + 2*βq*].

4*γq*² = *α*²*q*² − *αβq* + ¼*β* ² + 2*βq*³.

2*βq*³ + (*α*² − 4*γ*)*q*² − *αβq* + ¼*β* ² = 0, which is just a cubic in *q*, which is something we’re presumed to know how to solve. Using *q* we get *m* and *p* from *m*=−*β*/2*q* and *p*=½(*α*+*m*).

With these values of *m*, *p*, & *q*, *f* ≡ *g*. So the roots of *g*(*x*) from above are the roots of *f*(*x*), *et voilà*, we’re done, no magic needed.

^{*}A depressed *n*^{th} degree polynomial is one in which the (*n*−1)^{th} term has a coefficient of zero, so a depressed quartic is a quartic equation with no cubic term. For any *n*^{th} degree polynomial *F*(*x*) = *x ^{n}* +

*a*

_{n−1}

*x*

^{n−1}+

*a*

_{n−2}

*x*

^{n−2}+ …, there is a depressed

*n*

^{th}degree polynomial

*f*(

*x*) =

*x*+

^{n}*b*

_{n−2}

*x*

^{n−2}+ … such that

*F*(

*x*+

*s*) =

*f*(

*x*), because the coefficient of the (

*n*−1)

^{th}term of

*F*(

*x*+

*s*) is

*ns*+

*a*

_{n−1}, which can always be made zero by selecting

*s*to be –

*a*

_{n−1}/n.